An α-particle of kinetic energy \(K\) moves straight towards a nucleus of atomic number \(Z\). It stops for a moment at a distance \(d\) from the centre of the nucleus and then turns back. Which relation holds at this stopping point?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d}\)
The total mechanical energy of the α-particle and the nucleus stays the same during the collision. Far from the nucleus, this energy is all kinetic. At the stopping point the speed is zero, so the energy is all electric potential energy.
Option (1) \(K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d^{2}}\) — Incorrect. The term with \( d^{2} \) is the Coulomb force at the stopping point. Potential energy varies as \( 1/d \).
Option (2) \(K=\frac{1}{4\pi\varepsilon_0}\frac{(e)(Ze)}{d}\) — Incorrect. An α-particle is the nucleus of a helium atom. It has two units of positive charge, so its charge is \( 2e \).
Option (3) \(K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d}\) — Correct. The whole kinetic energy \( K \) has turned into the potential energy of the charges \( 2e \) and \( Ze \) at distance \( d \).
Option (4) \(K=\frac{1}{2}\cdot\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d}\) — Incorrect. There is no factor of one-half here. The particle comes fully to rest, so all of its kinetic energy becomes potential energy.
NCERT (2026-27):
Example 12.2 – “Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy Ei = Ef as”
\( \displaystyle K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d} \)\( \displaystyle =\frac{2Ze^{2}}{4\pi\varepsilon_0d} \)